A295053 Solution of the complementary equation a(n) = a(n-1) + a(n-2) + b(0) + b(1) + ... + b(n-1), where a(0) = 1, a(1) = 2, b(0) = 3, and (a(n)) and (b(n)) are increasing complementary sequences.

1, 2, 10, 24, 52, 101, 186, 329, 568, 962, 1608, 2662, 4377, 7162, 11679, 18999, 30855, 50051, 81124, 131415, 212802, 344505, 557621, 902467, 1460457, 2363322, 3824207, 6187988, 10012686, 16201198, 26214442, 42416233, 68631304, 111048203

Offset: 0

Author

Clark Kimberling, Nov 18 2017

Keywords

• nonn
• easy

Comments

The increasing complementary sequences a() and b() are uniquely determined by the titular equation and initial values. Guide to related sequencres:

A295053: a(n) = a(n-1) + a(n-2) + b(0) + b(1) + ... + b(n-1), a(0) = 1, a(1) = 2, b(0) = 3

A295054: a(n) = a(n-1) + a(n-2) + b(1) + b(2) + ... + b(n-1), a(0) = 1, a(1) = 2, b(0) = 3

A295055: a(n) = a(n-2) + b(1) + b(2) + ... + b(n-1), a(0) = 1, a(1) = 2, b(0) = 3

A295056: a(n) = 2*a(n-1) + b(n-1), a(0) = 1, a(1) = 4, b(0) = 2, b(1) = 3

A295057: a(n) = 2*a(n-1) + b(n-1), a(0) = 2, a(1) = 5, b(0) = 1

A295058: a(n) = 2*a(n-1) - b(n-1), a(0) = 3, a(1) = 5, b(0) = 1

A295059: a(n) = 2*a(n-1) + b(n-2), a(0) = 1, a(1) = 4, b(0) = 2, b(1) = 3

A295060: a(n) = 2*a(n-1) - b(n-2), a(0) = 3, a(1) = 5, b(0) = 1, b(1) = 2

A295061: a(n) = 4*a(n-1) + b(n-1), a(0) = 1, a(1) = 3, b(0) = 2

A295062: a(n) = 4*a(n-2) + b(n-2), a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4

A295063: a(n) = 4*a(n-2) + b(n-1) + b(n-2), a(0) = 1, a(1) = 3, b(0) = 2

A295064: a(n) = 8*a(n-3) + b(n-1), a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2

A295065: a(n) = 8*a(n-3) + b(n-2), a(0) = 1, a(1) = 3, a(2) = 5, b(0) = 2

A295066: a(n) = 2*a(n-2) + b(n-1), a(0) = 1, a(1) = 3, b(0) = 2

A295067: a(n) = 2*a(n-2) + b(n-2), a(0) = 1, a(1) = 3, b(0) = 2

A295068: a(n) = 2*a(n-2) - b(n-1) + n, a(0) = 3, a(1) = 4, b(0) = 1

A295069: a(n) = 2*a(n-2) - b(n-2) + n, a(0) = 3, a(1) = 4, b(0) = 1

A295070: a(n) = a(n-2) + b(n-1) + b(n-2), a(0) = 3, a(1) = 2, b(0) = 3

A295133: a(n) = 3*a(n-1) + b(n-1), a(0) = 1, a(1) = 2, b(0) = 3

A295134: a(n) = 3*a(n-1) + b(n-1) - 1, a(0) = 1, a(1) = 2, b(0) = 3

A295135: a(n) = 3*a(n-1) + b(n-1) - 2, a(0) = 1, a(1) = 2, b(0) = 3

A295136: a(n) = 3*a(n-1) + b(n-1) - 3, a(0) = 1, a(1) = 2, b(0) = 3

A295137: a(n) = 3*a(n-1) + b(n-1) - n, a(0) = 1, a(1) = 2, b(0) = 3

A295138: a(n) = 3*a(n-2) + b(n-1), a(0) = 1, a(1) = 2, b(0) = 3

A295139: a(n) = 3*a(n-1) + b(n-2), a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4

A295140: a(n) = 3*a(n-1) - b(n-2) + 4, a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4

A295141: a(n) = 2*a(n-1) + a(n-2) + b(n-2), a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4

A295142: a(n) = 2*a(n-1) + a(n-2) + b(n-2), a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4

A295143: a(n) = 2*a(n-1) + a(n-1) + b(n-1), a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4

A295144: a(n) = 2*a(n-1) + a(n-2) + b(n-1), a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4

A295145: a(n) = a(n-1) + 2*a(n-2) + b(n-2), a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4

A295146: a(n) = a(n-1) + 2*a(n-2) + b(n-2), a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4

A295147: a(n) = a(n-1) + 2*a(n-2) + b(n-1), a(0) = 1, a(1) = 2, b(0) = 3, b(1) = 4

A295148: a(n) = a(n-1) + 2*a(n-2) + b(n-1), a(0) = 1, a(1) = 3, b(0) = 2, b(1) = 4

Examples

			a(0) = 1, a(1) = 2, b(0) = 3
b(1) = 4 (least "new number")
a(2) = a(1) + a(0) + b(0) + b(1) = 10
Complement: (b(n)) = (3, 4, 5, 6, 7, 8, 9, 11, 12, 13, 14, ...)
		

Links

• Clark Kimberling, Complementary equations, J. Int. Seq. 19 (2007), 1-13.

Crossrefs

Cf. A294860.

Programs

Mathematica

mex := First[Complement[Range[1, Max[#1] + 1], #1]] &;
a[0] = 1; a[1] = 2; b[0] = 3;
a[n_] := a[n] = a[n - 1] + a[n - 2] + Sum[b[k], {k, 0, n - 1}];
b[n_] := b[n] = mex[Flatten[Table[Join[{a[n]}, {a[i], b[i]}], {i, 0, n - 1}]]];
Table[a[n], {n, 0, 18}]  (* A295053 *)
Table[b[n], {n, 0, 10}]